QJEEM24Ph6167: JEE Main 2024 Physics - Class 12 JEEM24Ph6167 Solution: The coercivity of a magnet is 5 × 10 3   A / m . The…

Coercivity is a measure of the resistance of a ferromagnetic material to becoming demagnetized. It is defined as the intensity of the applied magnetic field required to reduce the magnetization of a material to zero after the magnetization of the sample has been driven to saturation. In this case, coercivity $$H_c$$ is given to be $$5\times {10}^3\mathrm{A}/\mathrm{m}$$.

The relationship between the magnetic field $$H$$ inside a solenoid and the current $$I$$ passed through it is given by the formula:

$$H=\frac{N\cdot I}{L}$$

where:

• $$H$$ is the magnetic field strength inside the solenoid in amperes per meter (A/m),
• $$N$$ is the total number of turns of wire,
• $$I$$ is the current in amperes (A), and
• $$L$$ is the length of the solenoid in meters (m).

Given that the number of turns of the solenoid $$N=150$$ and the length of the solenoid $$L=30\text{cm}=0.3\text{m}$$, we can rearrange the formula to solve for $$I$$:

$$I=\frac{H\cdot L}{N}$$

Substituting the given values:

$$I=\frac{(5\times {10}^3)\times 0.3}{150}$$

Simplifying, we get:

$$I=\frac{1500}{150}$$

$$I=10\text{A}$$

Therefore, the amount of current required to be passed in the solenoid for demagnetizing the magnet when inside the solenoid is 10 A.